A1.25-g sample contains some of the very reactive compound al(c6h5)3. on treating the compound with aqueous hcl, 0.951 g of c6h6 is obtained. al(c6h5)3(s) + 3hcl(aq) ® alcl3(aq) + 3c6h6(l) assuming that al(c6h5)3 was converted completely to products, what is the weight percent of al(c6h5)3 in original 1.25-g sample?

83.9% is the weight percent of Al(C6H5)3 in the original 1.25 g sample. First, look up the atomic weights of all elements involved. Atomic weight of Aluminum = 26.981539 Atomic weight of Carbon = 12.0107 Atomic weight of Chlorine = 35.453 Atomic weight of Hydrogen = 1.00794 Now calculate the molar mass of Al(C6H5)3, and C6H6 Molar mass Al(C6H5)3 = 26.981539 + 18 * 12.0107 + 15 * 15.999 = 258.293239 g/mol Molar mass of C6H6 = 6 * 12.0107 + 6 * 1.00794 = 78.11184 g/mol Determine how many moles of C6H6 was produced 0.951 g / 78.11184 g/mol = 0.012174851 mol Since the balanced formula indicates that 3 moles of C6H6 is produced for each mole of Al(C6H5)3 used, divide by 3 to get the number of moles of Al(C6H5)3 that was present. 0.012174851 mol / 3 = 0.004058284 mol Now multiply by the molar mass of Al(C6H5)3 to get the mass of Al(C6H5)3 originally present. 0.004058284 mol * 258.293239 g/mol = 1.048227218 g Finally, divide the mass of Al(C6H5)3 by the total mass of the original sample to get the weight percentage. 1.048227218 g / 1.25 g = 0.838582 Since all our measurements had 3 significant figures, round the result to 3 significant figures, giving 0.839 = 83.9%

The answer is b because a,c,and d are opinons